Radio active decay

Q. The half life of carbon-14 is 5730 years. If the original amount of carbon-14 in a particular living organism is 20 grams and that found in a fossil is 6 grams. determine the approximate age of the fossil.

You know, the rate decreases proportional to the amount present. Solving this we get:

[tex]A(t)=A_0e^{-kt}[/tex]

Now, the half-life is related to the decay constant k as follows:

[tex]t_{1/2}=\frac{ln(2)}{k}[/tex]

So, you can plug this into the equation, supply the amounts for the half-life and the initial amount and what's left, do some algebra, you can do that, and then solve for the time when only 6 grams are left.

Did I give that one away? You know what, you need to figure out what the decay constant is. Ignore that man behind the curtain . . . do what Daniel said.

The way I would do this problem is: If the half life is 5730 years, then the amount left, of C initially, after t years, is [tex]A(t)= C\left( \frac{1}{2}\right)^{\frac{t}{5730}}[/tex]. (Notice that t/5730 just measures the number of times the quantity is "halved".)

You are told that C= 20 and that A(t)= 6. Solve [tex]20\left(\frac{1}{2}\right)^{\frac{t}{5730}}= 6[/tex] for t.

Naeem, let's think about this for a while: The half-life is 5730 and you start with 20 grams. So after 5730 years you'll have 10 grams, and after 11460 years you'll have 5 grams. So at 12039 years you'll have less than 5 grams.

I was presented with a radioactivity problem which does not make any sense to me. Hope it does to someone in here:

A radioactive element releasing 8 rays (whatever this means) is giving a Geiger count of 188 per minute at the beginning of a test. 40 minutes down the line the count drops to 53. What is the half life??? (The answer is apparently 20 but HOW?)

I dunno where that 8 comes into turn.I think they might mean counting the rays.Which initially (in the first minute) would be 188 and after 40mins it would drop to 53,which is just what my eq.stated...